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Value of a Polynomial at x = n+1 (Tomato Subjective 176)

Suppose that P(x) is a polynomial of degree n such that \(P(k) = \frac {k}{k+1} \) for k = 0, 1, 2, …, n . Find the value of P(n+1). (Subjective Problem 176 from Test of Mathematics)

Discussion:

Consider an auxiliary polynomial g(x) = (x+1)P(x) – x . g(x) is an n+1 degree polynomial (as P(x) is n degree and we multiply (x+1) with it). We note that g(0) = g(1) = … = g(n) = 0  (as the given condition allows (k+1) P(k) – k = 0 for all k from 0 to n). Hence 0, 1, 2, … , n are the n+1 roots of g(x).

 

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